Difference between revisions of "Aufgaben:Exercise 5.1: Error Distance Distribution"

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{{quiz-Header|Buchseite=Digitalsignalübertragung/Beschreibungsgrößen digitaler Kanalmodelle}}
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{{quiz-Header|Buchseite=Digital_Signal_Transmission/Parameters_of_Digital_Channel_Models}}
  
[[File:P_ID1827__Dig_A_5_1.png|right|frame|Gegebene Verteilung der Fehlerabstände]]
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[[File:EN_Dig_A_5_1.png|right|frame|Error distance distribution]]
Ein jedes digitales Kanalmodell kann in gleicher Weise beschrieben werden durch
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Any digital channel model can be described in the same way by
* die Fehlerfolge $〈e_{\rm \nu}〉$,
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* the error sequence  $〈e_{\rm \nu}〉$, and
* durch die Fehlerabstandsfolge $〈a_{\rm \nu '}〉$.
 
  
 +
* the error distance sequence  $〈a_{\rm \nu \hspace{0.05cm}'}〉$.
  
Beispielhaft betrachten wir die Folgen:
+
 
 +
As an example,  we consider the sequences:
 
:$$<\hspace{-0.1cm}e_{\nu} \hspace{-0.1cm}>  \ = \ <
 
:$$<\hspace{-0.1cm}e_{\nu} \hspace{-0.1cm}>  \ = \ <
 
\hspace{-0.1cm}0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 0, 1, \text{...}
 
\hspace{-0.1cm}0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 0, 1, \text{...}
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\hspace{-0.1cm}> \hspace{0.05cm}.$$
 
\hspace{-0.1cm}> \hspace{0.05cm}.$$
  
Man erkennt daraus beispielsweise:
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One can see from this,&nbsp; for example:
* Der Fehlerabstand $a_2 = 3$ bedeutet, dass zwischen dem ersten und dem zweiten Fehler zwei fehlerfreie Symbole liegen.
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* The error distance&nbsp; $a_2 = 3$&nbsp; means that there are two error-free symbols between the first and the second error.
* Dagegen  deutet $a_3 = 1$ darauf hin, dass nach dem zweiten Fehler direkt ein dritter folgt.
 
  
 +
* In contrast,&nbsp; $a_3 = 1$&nbsp; indicates that the second error is immediately followed by a third.
  
Die unterschiedlichen Laufindizes ($\nu$ und $\nu\hspace{0.05cm} '$, jeweils beginnend mit $1$) sind erforderlich, da keine Synchronität zwischen der Fehlerabstandsfolge und der Fehlerfolge besteht.
 
  
In der Grafik ist für zwei verschiedene Modelle $M_1$ und $M_2$ die Fehlerabstandsverteilung (FAV)  
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The different indices &nbsp;$(\nu$&nbsp; and&nbsp; $\nu\hspace{0.05cm} '$,&nbsp; each starting with &nbsp;$1$)&nbsp; are necessary because there is no synchrony between the error distance sequence and the error sequence.
 +
 
 +
In the graph,&nbsp; for two different models &nbsp;$M_1$&nbsp; and &nbsp;$M_2$,&nbsp; the&nbsp; "error distance distribution"&nbsp; $\rm (EDD)$&nbsp; is given as
 
:$$V_a(k) =  {\rm Pr}(a \ge k) = 1 - \sum_{\kappa = 1}^{k}  {\rm Pr}(a = \kappa)\hspace{0.05cm}$$
 
:$$V_a(k) =  {\rm Pr}(a \ge k) = 1 - \sum_{\kappa = 1}^{k}  {\rm Pr}(a = \kappa)\hspace{0.05cm}$$
  
angegeben. Diese Tabelle soll in dieser Aufgabe ausgewertet werden.
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This table is to be evaluated in this exercise.
  
  
  
''Hinweise:''
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* Die Aufgabe gehört zum Kapitel [[Digitalsignal%C3%BCbertragung/Beschreibungsgr%C3%B6%C3%9Fen_digitaler_Kanalmodelle| Beschreibungsgrößen digitaler Kanalmodelle]].
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Note:&nbsp;  The exercise belongs to the chapter&nbsp; [[Digital_Signal_Transmission/Parameters_of_Digital_Channel_Models|"Parameters of Digital Channel Models"]].
 
   
 
   
  
  
  
===Fragebogen===
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===Questions===
 
<quiz display=simple>
 
<quiz display=simple>
{Wie lauten die folgenden Fehlerwerte ($0$ oder $1$)?
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{What are the following error values &nbsp;$(0$&nbsp; or&nbsp; $1)$?
 
|type="{}"}
 
|type="{}"}
 
$e_{\rm 16} \ = \ $ { 0. }  
 
$e_{\rm 16} \ = \ $ { 0. }  
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$e_{\rm 18} \ = \ $ { 1 }  
 
$e_{\rm 18} \ = \ $ { 1 }  
  
{Wie groß ist bei beiden Modellen der Wert $V_a(k = 1)$?
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{What is the value of&nbsp; $V_a(k = 1)$ for both models?
 
|type="{}"}
 
|type="{}"}
 
$V_a(k = 1) \ = \ $ { 1 }  
 
$V_a(k = 1) \ = \ $ { 1 }  
  
{Bestimmen Sie für das Modell $M_1$ die Wahrscheinlichkeiten der Fehlerabstände.
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{For model&nbsp; $M_1$,&nbsp;&nbsp; determine the probabilities of the error distances.
 
|type="{}"}
 
|type="{}"}
 
${\rm Pr}(a = 1) \ = \ $ { 0.3 3% }  
 
${\rm Pr}(a = 1) \ = \ $ { 0.3 3% }  
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${\rm Pr}(a = 5) \ = \ $ { 0.1 3% }
 
${\rm Pr}(a = 5) \ = \ $ { 0.1 3% }
  
{Wie groß ist der maximal mögliche Fehlerabstand beim Modell $M_1$?
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{What is the maximum possible error distance for model&nbsp; $M_1$?
 
|type="{}"}
 
|type="{}"}
 
$k_{\rm max} \ = \ ${ 5 }  
 
$k_{\rm max} \ = \ ${ 5 }  
  
{Berechnen Sie für das Modell $M_1$ den mittleren Fehlerabstand.
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{Calculate the average error distance for model&nbsp; $M_1$.&nbsp;
 
|type="{}"}
 
|type="{}"}
${\rm E}[a] \ = \ ${ 2.5 3% }
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${\rm E}\big[a \big] \ = \ ${ 2.5 3% }
  
{Wie groß ist  beim Modell $M_1$ die mittlere Fehlerwahrscheinlichkeit $p_{\rm M} = {\rm E}[e]$?
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{For model&nbsp; $M_1$,&nbsp; what is the mean error probability&nbsp; $p_{\rm M} = {\rm E}[e]$?
 
|type="{}"}
 
|type="{}"}
 
$p_{\rm M} \ = \ ${ 0.4 3% }
 
$p_{\rm M} \ = \ ${ 0.4 3% }
  
{Welche Aussagen stimmen für das Modell $M_2$ mit Sicherheit?
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{Which statements are true for the model&nbsp; $M_2$&nbsp; with certainty?
 
|type="[]"}
 
|type="[]"}
+ Zwei Fehler können nicht direkt aufeinander folgen.
+
+ Two errors cannot directly follow each other.
- Der häufigste Fehlerabstand ist $a = 6$.
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- The most frequent error distance is&nbsp; $a = 6$.
- Die mittlere Fehlerwahrscheinlichkeit beträgt $p_{\rm M} = 0.25$.
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- The mean error probability is&nbsp; $p_{\rm M} = 0.25$.
 
</quiz>
 
</quiz>
  
===Musterlösung===
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===Solution===
 
{{ML-Kopf}}
 
{{ML-Kopf}}
'''(1)'''&nbsp; Die Auswertung der Fehlerabstandsfolge weist auf Fehler bei $\nu = 2, 5, 6, 10, 12, 17, 18, 19, 22, 26, 27$ und $29$ hin. Daraus folgt:
+
'''(1)'''&nbsp; Evaluation of the error distance sequence indicates errors at&nbsp; $\nu = 2, 5, 6, 10, 12, 17, 18, 19, 22, 26, 27$&nbsp; and&nbsp; $29$.
* $e_{\rm 16} \ \underline {= 0}$,
+
* $e_{\rm 17} \ \underline {= 1}$,
+
*It follows: &nbsp; $e_{\rm 16} \ \underline {= 0}$, &nbsp; &nbsp; $e_{\rm 17} \ \underline {= 1}$, &nbsp; &nbsp; $e_{\rm 18} \ \underline {= 1}$.
* $e_{\rm 18} \ \underline {= 1}$.
 
  
  
'''(2)'''&nbsp; Aus der Definitionsgleichung folgt bereits
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'''(2)'''&nbsp; From the definition equation follows already
 
:$$V_a(k = 1) =  {\rm Pr}(a \ge 1)\hspace{0.15cm}\underline {= 1} \hspace{0.05cm}.$$
 
:$$V_a(k = 1) =  {\rm Pr}(a \ge 1)\hspace{0.15cm}\underline {= 1} \hspace{0.05cm}.$$
  
'''(3)'''&nbsp; Es gilt ${\rm Pr}(a = k) = V_a(k) \, &ndash;V_a(k+1)$. Daraus erhält man für die einzelnen Wahrscheinlichkeiten:
+
 
 +
'''(3)'''&nbsp; ${\rm Pr}(a = k) = V_a(k) \, &ndash;V_a(k+1)$&nbsp; holds.&nbsp; From this we obtain for the individual probabilities:
 
:$${\rm Pr}(a = 1)\hspace{-0.1cm} \ = \ \hspace{-0.1cm}V_a(1) - V_a(2) = 1 - 0.7\hspace{0.15cm}\underline {= 0.3}\hspace{0.05cm},$$
 
:$${\rm Pr}(a = 1)\hspace{-0.1cm} \ = \ \hspace{-0.1cm}V_a(1) - V_a(2) = 1 - 0.7\hspace{0.15cm}\underline {= 0.3}\hspace{0.05cm},$$
 
:$${\rm Pr}(a = 2)\hspace{-0.1cm} \ = \ \hspace{-0.1cm}V_a(2) - V_a(3) = 0.7 - 0.45 \hspace{0.15cm}\underline {= 0.25}\hspace{0.05cm},$$
 
:$${\rm Pr}(a = 2)\hspace{-0.1cm} \ = \ \hspace{-0.1cm}V_a(2) - V_a(3) = 0.7 - 0.45 \hspace{0.15cm}\underline {= 0.25}\hspace{0.05cm},$$
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V_a(6) = 0.10 - 0 \hspace{0.15cm}\underline {= 0.10}\hspace{0.05cm}.$$
 
V_a(6) = 0.10 - 0 \hspace{0.15cm}\underline {= 0.10}\hspace{0.05cm}.$$
  
'''(4)'''&nbsp; Aus $V_a(k=6) = {\rm Pr}(a &#8805; 6) = 0$ folgt für den maximalen Fehlerabstand direkt $k_{\rm max} \ \underline {= 5}$.
+
 
 +
'''(4)'''&nbsp; From&nbsp; $V_a(k=6) = {\rm Pr}(a &#8805; 6) = 0$,&nbsp; it follows directly for the maximum error distance&nbsp;
 +
:$$k_{\rm max} \ \underline {= 5}.$$
  
  
'''(5)'''&nbsp; Mit den unter (3) berechneten Wahrscheinlichkeiten ergibt sich für den gesuchten Erwartungswert:
+
'''(5)'''&nbsp; Using the probabilities calculated in subtask&nbsp; '''(3)''',&nbsp; the expected value we are looking for is:
:$${\rm E}[a] = \sum_{k = 1}^{5} k \cdot {\rm Pr}(a = k) =  1 \cdot 0.3 +2 \cdot 0.25 +3 \cdot 0.2 +4 \cdot 0.15 +5 \cdot 0.1\hspace{0.15cm}\underline { = 2.5}
+
:$${\rm E}\big[a \big] = \sum_{k = 1}^{5} k \cdot {\rm Pr}(a = k) =  1 \cdot 0.3 +2 \cdot 0.25 +3 \cdot 0.2 +4 \cdot 0.15 +5 \cdot 0.1\hspace{0.15cm}\underline { = 2.5}
 
  \hspace{0.05cm}.$$
 
  \hspace{0.05cm}.$$
  
'''(6)'''&nbsp; Die mittlere Fehlerwahrscheinlichkeit ist der Kehrwert des mittleren Fehlerabstands: $p_{\rm M} \ \underline {= 0.4}$.
 
  
 +
'''(6)'''&nbsp; The mean error probability is the inverse of the average error distance: &nbsp;
 +
:$$p_{\rm M} \ \underline {= 0.4}.$$
 +
 +
 +
'''(7)'''&nbsp; With certainty,&nbsp; only <u>statement 1</u>&nbsp; is true:
 +
*The first statement is true because ${\rm Pr}(a = 1) = V_a(1) - V_a(2) = 0$.
 +
 +
* The second statement is not certain because&nbsp; $V_a(6)$&nbsp; gives only the sum of the probabilities ${\rm Pr}(a &#8805; 6)$,&nbsp; but not ${\rm Pr}(a = 6)$ alone.&nbsp;
 +
 +
*Only with the additional specification&nbsp; $V_a(7) = 0$&nbsp; would statement 2 be true.
  
'''(7)'''&nbsp; Mit Sicherheit stimmt nur die <u>Aussage 1</u>:
+
* Likewise,&nbsp; for the expected value&nbsp; ${\rm E}[a]$,&nbsp; no definite statement is possible due to missing information.&nbsp; With $V_a(7) = 0$ the result would be:
*Die Aussage 1 stimmt, da ${\rm Pr}(a = 1) = V_a(1) \, &ndash; V_a(2) = 0$ ist.
 
* Die zweite Aussage ist nicht sicher, da $V_a(6)$ nur die Summe der Wahrscheinlichkeiten ${\rm Pr}(a &#8805; 6)$ angibt, aber nicht ${\rm Pr}(a = 6)$ allein. Nur mit der zusätzlichen Angabe $V_a(7) = 0$ würde die Aussage 2 zutreffen.
 
* Ebenso ist für den Erwartungswert ${\rm E}[a]$ augrund fehlender Angaben keine endgültige Aussage möglich. Mit $V_a(7 = 0)$ würde sich ergeben.:
 
 
:$${\rm E}[a] =  2 \cdot 0.1 +3 \cdot 0.2 +4 \cdot 0.2 +5 \cdot 0.2 +6 \cdot 0.3=
 
:$${\rm E}[a] =  2 \cdot 0.1 +3 \cdot 0.2 +4 \cdot 0.2 +5 \cdot 0.2 +6 \cdot 0.3=
  4.4$$
+
  4.4.$$
*Ohne diese Angabe ist nur die Aussage ${\rm E}[a] &#8805; 4.4$ möglich. Damit gilt aber für die mittlere Fehlerwahrscheinlichkeit die Bedingung $p_{\rm M} < 1/4.4 < 0.227$ &nbsp; &#8658; &nbsp; Die Aussage 3 trifft also mit Sicherheit nicht zu. .
+
*Without this specification,&nbsp; only the statement&nbsp; ${\rm E}[a] &#8805; 4.4$&nbsp; is possible. But this means that the condition&nbsp; $p_{\rm M} < 1/4.4 < 0.227$&nbsp; is valid for the mean error probability.&nbsp; Statement 3 is therefore also not true with certainty.
 
{{ML-Fuß}}
 
{{ML-Fuß}}
  
  
[[Category:Aufgaben zu Digitalsignalübertragung|^5.1 Zu den Digitalen Kanalmodellen^]]
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[[Category:Digital Signal Transmission: Exercises|^5.1 Digital Channel Models^]]

Latest revision as of 14:13, 1 October 2022

Error distance distribution

Any digital channel model can be described in the same way by

  • the error sequence  $〈e_{\rm \nu}〉$, and
  • the error distance sequence  $〈a_{\rm \nu \hspace{0.05cm}'}〉$.


As an example,  we consider the sequences:

$$<\hspace{-0.1cm}e_{\nu} \hspace{-0.1cm}> \ = \ < \hspace{-0.1cm}0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 0, 1, \text{...} \hspace{-0.1cm}> \hspace{0.05cm},$$
$$< \hspace{-0.1cm}a_{\nu\hspace{0.05cm} '} \hspace{-0.15cm}> \ = \ <\hspace{-0.1cm}2, 3, 1, 4, 2, 5, 1, 1, 3, 4, 1, 2, \text{...} \hspace{-0.1cm}> \hspace{0.05cm}.$$

One can see from this,  for example:

  • The error distance  $a_2 = 3$  means that there are two error-free symbols between the first and the second error.
  • In contrast,  $a_3 = 1$  indicates that the second error is immediately followed by a third.


The different indices  $(\nu$  and  $\nu\hspace{0.05cm} '$,  each starting with  $1$)  are necessary because there is no synchrony between the error distance sequence and the error sequence.

In the graph,  for two different models  $M_1$  and  $M_2$,  the  "error distance distribution"  $\rm (EDD)$  is given as

$$V_a(k) = {\rm Pr}(a \ge k) = 1 - \sum_{\kappa = 1}^{k} {\rm Pr}(a = \kappa)\hspace{0.05cm}$$

This table is to be evaluated in this exercise.



Note:  The exercise belongs to the chapter  "Parameters of Digital Channel Models".



Questions

1

What are the following error values  $(0$  or  $1)$?

$e_{\rm 16} \ = \ $

$e_{\rm 17} \ = \ $

$e_{\rm 18} \ = \ $

2

What is the value of  $V_a(k = 1)$ for both models?

$V_a(k = 1) \ = \ $

3

For model  $M_1$,   determine the probabilities of the error distances.

${\rm Pr}(a = 1) \ = \ $

${\rm Pr}(a = 2) \ = \ $

${\rm Pr}(a = 3) \ = \ $

${\rm Pr}(a = 4) \ = \ $

${\rm Pr}(a = 5) \ = \ $

4

What is the maximum possible error distance for model  $M_1$?

$k_{\rm max} \ = \ $

5

Calculate the average error distance for model  $M_1$. 

${\rm E}\big[a \big] \ = \ $

6

For model  $M_1$,  what is the mean error probability  $p_{\rm M} = {\rm E}[e]$?

$p_{\rm M} \ = \ $

7

Which statements are true for the model  $M_2$  with certainty?

Two errors cannot directly follow each other.
The most frequent error distance is  $a = 6$.
The mean error probability is  $p_{\rm M} = 0.25$.


Solution

(1)  Evaluation of the error distance sequence indicates errors at  $\nu = 2, 5, 6, 10, 12, 17, 18, 19, 22, 26, 27$  and  $29$.

  • It follows:   $e_{\rm 16} \ \underline {= 0}$,     $e_{\rm 17} \ \underline {= 1}$,     $e_{\rm 18} \ \underline {= 1}$.


(2)  From the definition equation follows already

$$V_a(k = 1) = {\rm Pr}(a \ge 1)\hspace{0.15cm}\underline {= 1} \hspace{0.05cm}.$$


(3)  ${\rm Pr}(a = k) = V_a(k) \, –V_a(k+1)$  holds.  From this we obtain for the individual probabilities:

$${\rm Pr}(a = 1)\hspace{-0.1cm} \ = \ \hspace{-0.1cm}V_a(1) - V_a(2) = 1 - 0.7\hspace{0.15cm}\underline {= 0.3}\hspace{0.05cm},$$
$${\rm Pr}(a = 2)\hspace{-0.1cm} \ = \ \hspace{-0.1cm}V_a(2) - V_a(3) = 0.7 - 0.45 \hspace{0.15cm}\underline {= 0.25}\hspace{0.05cm},$$
$${\rm Pr}(a = 3)\hspace{-0.1cm} \ = \ \hspace{-0.1cm}V_a(3) - V_a(4) = 0.45 - 0.25 \hspace{0.15cm}\underline {= 0.2}\hspace{0.05cm},$$
$${\rm Pr}(a = 4)\hspace{-0.1cm} \ = \ \hspace{-0.1cm}V_a(4) - V_a(5) = 0.25 - 0.10 \hspace{0.15cm}\underline {= 0.15}\hspace{0.05cm},$$
$${\rm Pr}(a = 5)\hspace{-0.1cm} \ = \ \hspace{-0.1cm}V_a(5) - V_a(6) = 0.10 - 0 \hspace{0.15cm}\underline {= 0.10}\hspace{0.05cm}.$$


(4)  From  $V_a(k=6) = {\rm Pr}(a ≥ 6) = 0$,  it follows directly for the maximum error distance 

$$k_{\rm max} \ \underline {= 5}.$$


(5)  Using the probabilities calculated in subtask  (3),  the expected value we are looking for is:

$${\rm E}\big[a \big] = \sum_{k = 1}^{5} k \cdot {\rm Pr}(a = k) = 1 \cdot 0.3 +2 \cdot 0.25 +3 \cdot 0.2 +4 \cdot 0.15 +5 \cdot 0.1\hspace{0.15cm}\underline { = 2.5} \hspace{0.05cm}.$$


(6)  The mean error probability is the inverse of the average error distance:  

$$p_{\rm M} \ \underline {= 0.4}.$$


(7)  With certainty,  only statement 1  is true:

  • The first statement is true because ${\rm Pr}(a = 1) = V_a(1) - V_a(2) = 0$.
  • The second statement is not certain because  $V_a(6)$  gives only the sum of the probabilities ${\rm Pr}(a ≥ 6)$,  but not ${\rm Pr}(a = 6)$ alone. 
  • Only with the additional specification  $V_a(7) = 0$  would statement 2 be true.
  • Likewise,  for the expected value  ${\rm E}[a]$,  no definite statement is possible due to missing information.  With $V_a(7) = 0$ the result would be:
$${\rm E}[a] = 2 \cdot 0.1 +3 \cdot 0.2 +4 \cdot 0.2 +5 \cdot 0.2 +6 \cdot 0.3= 4.4.$$
  • Without this specification,  only the statement  ${\rm E}[a] ≥ 4.4$  is possible. But this means that the condition  $p_{\rm M} < 1/4.4 < 0.227$  is valid for the mean error probability.  Statement 3 is therefore also not true with certainty.